If two straight lines whose direction cosines are given by the relations $l+m-n=0,3l^2+m^2+cnl=0$ are parallel, then the positive value of $c$ is 

The given relations are $l+m-n=0,3l^2+m^2+cnl=0$

Suppose that $l=1$

Hence, the equations are $m=-1+n$

substitute the above values in the equation $3l^2+m^2+cnl=0$

$\begin{array}{rcl}3+{\left(-1+n\right)}^2+cn& =& 0\\ n^2-2n+1+3+cn& =& 0\\ n^2+\left(-2+c\right)n+4& =& 0\end{array}$

Since the lines are parallel, the above equation has equal real roots

Hence, 

  $\begin{array}{rcl}{\left(-2+c\right)}^2-16& =& 0\\ -2+c& =& \pm 4\\ c& =& 6,-2\end{array}$

The positive value of $c$ is $c=\boxed{6}$